Question

Two blocks, A and B, of the masses 3,3kg and 2,1kg respectively, are joined by an inextensible string. The string is threaded through a frictionless pulley allowing block B to be suspended by it. The coefficient between block A and the surface of the table is 0,1.
Figure 6: Diagram showing two blocks connected by string
Calculate the acceleration of the blocks.

Answer 1

To calculate the acceleration of the blocks, we first need to consider the forces acting on each block.

For block A:
- The weight of block A is given by: F1 = m1 * g = 3,3kg * 9,8m/s^2 = 32,34N
- The normal force acting on block A is equal in magnitude and opposite in direction to the weight, so: N1 = 32,34N
- The frictional force acting on block A is given by: f = µ * N1 = 0,1 * 32,34N = 3,234N
- The net force acting on block A is given by: Fnet = T - f = m1 * a, where T is the tension in the string and a is the acceleration of the blocks

For block B:
- The weight of block B is given by: F2 = m2 * g = 2,1kg * 9,8m/s^2 = 20,58N
- The tension in the string is equal in magnitude and opposite in direction to the weight, so: T = 20,58N

Since block B is being lifted, the direction of acceleration will be upwards. Let's consider the upwards direction as positive.

For block B:
- The net force acting on block B is given by: T - m2 * g = m2 * a = 2,1kg * a

Using the information above, we can equate the net forces for both blocks and solve for the acceleration:

m1 * a = T - f
3,3kg * a = 20,58N - 3,234N
3,3kg * a = 17,346N
a = 5,26m/s^2

Therefore, the acceleration of the blocks is 5,26m/s^2.